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Stimulated by some Physics backgrounds, consider the following two sets of matrices.

Notations and definitions:Let $A,B$ be two complex $n\times n$ matrices, then $\left [ A,B \right ]\overset{\underset{\mathrm{def}}{}}{=}AB-BA$ and $ e^A\overset{\underset{\mathrm{def}}{}}{=} \sum_{m=0}^{\infty }\frac{A^m}{m!}. $

Let $M_1,M_2,M_3$ be three $n\times n$ Hermitian matrices, and they satisfy $[M_1,M_2]=iM_3,[M_2,M_3]=iM_1,[M_3,M_1]=iM_2$ identities, where $i=\sqrt{-1}$.

Now define a set $X$ of unitary matrices:$X=\left \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ and another set $Y$ of unitary matrices:$Y=\left \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ where $i=\sqrt{-1}$ ( Note that the number indices of the Hermitian matrices $M$ are different in $X$ and $Y$ ).

And my questions are as follows:

(1) Is $X=Y$ ?

(2) If (1) is true, is the set $X$ a group ?

(3) If both (1) and (2) are true, is $X\cong SU(2)$ true ?

Supplements: For concreteness, let's take a look at the following simple example. Consider the Physically called spin-$\frac{1}{2}$ "Pauli" matrices $M_1=\frac{1}{2}\bigl(\begin{smallmatrix} 0& 1\\ 1&0 \end{smallmatrix}\bigr),M_2=\frac{1}{2}\bigl(\begin{smallmatrix} 0& -i\\ i&0 \end{smallmatrix}\bigr)$ and $M_3=\frac{1}{2}\bigl(\begin{smallmatrix} 1& 0\\ 0&-1 \end{smallmatrix}\bigr),$ and it's easy to find that $M_1^2+M_2^2+M_3^2=\frac{1}{2}(\frac{1}{2}+1)\mathbb{I}$, where $\mathbb{I}$ is a $2\times2$ identity matrix.

In the above example, direct calculation of matrices $e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}$ in $X$ shows that $X=SU(2)$ (then $X$ is a group), and it's also easy to verify that $Y\subseteq SU(2)$. So now the question is, is $SU(2)\subseteq Y$ too ?

Thanks in advance.

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    $\begingroup$ In the definition of X there is $M_3$ twice. Is this an error or not? :) $\endgroup$
    – Riccardo
    May 3, 2013 at 21:06
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    $\begingroup$ I would imagine that any answer would depend heavily on $M_1, M_2$ and $M_3$, which you have not elaborated on. Could be wrong though. I don't immediately grasp what those commutation conditions say. $\endgroup$
    – rschwieb
    May 3, 2013 at 21:20
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    $\begingroup$ @ Ric Ped, Dear Ped, $M_3$ definitely appears twice, it's not an error. $\endgroup$
    – Kai Li
    May 3, 2013 at 21:33
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    $\begingroup$ @ Ric Ped, in fact, if you choose $M_1=\bigl(\begin{smallmatrix} 0 &1 \\ 1&0 \end{smallmatrix}\bigr),M_2=\bigl(\begin{smallmatrix} 0 &-i \\ i&0 \end{smallmatrix}\bigr),M_3=\bigl(\begin{smallmatrix} 1 &0 \\ 0&-1 \end{smallmatrix}\bigr)$, then $X$ is just the Euler's angles $(\alpha,\beta ,\gamma) $ representation of group $SU(2).$ $\endgroup$
    – Kai Li
    May 3, 2013 at 21:39
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    $\begingroup$ Those relations don't hold for the Euler's angles matrices, instead $[M_1,M_2]=2iM_3$ $[M_2,M_3]=2iM_1$ $[M_3,M_1]=2iM_2$. Also for those matrices $\sum M_i^2=3\mathbb{I}$, where you say it should be $\frac{3}{4}\mathbb{I}$. Are you missing a factor of $2$ somewhere? $\endgroup$
    – Alexander Gruber
    May 3, 2013 at 22:48

1 Answer 1

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Comments to the question (v3):

(We will from now on assume that the $n\times n$ matrices $M_1$, $M_2$ and $M_3$ are linearly independent, and that $n\geq 2$.)

OP defines two sets:

$$\tag{1} X_n~:=~ \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}\in {\rm Mat}_{n\times n}(\mathbb{C}) \mid \alpha,\beta,\gamma \in \mathbb{R} \}, $$

$$\tag{2} Y_n~:=~ \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} \in {\rm Mat}_{n\times n}(\mathbb{C}) \mid\alpha,\beta,\gamma \in \mathbb{R} \}.$$

The set $Y_n$ is the image $Y_n=\rho(SU(2))$ of an $n$-dimensional $SU(2)$ group representation $\rho:SU(2) \to GL(n,\mathbb{C})$, also known in physics as a spin $\frac{n-1}{2}$ representation. In particular, the set $Y$ is itself a group.

$$\tag{3} Y_n~\cong~\left\{ \begin{array}{rcl} SU(2)/ \mathbb{Z}_2 ~\cong~SO(3) &\text{for}&n&\text{odd}, \\ SU(2) &\text{for}&n&\text{even}.\end{array} \right. $$

The set $X_n\subseteq Y_n$ is a subset of $Y_n$; due, among other things, to the Baker-Campbell-Hausdorff formula.

In fact, the set $X_n$ is a generalized Euler angle realization of $SU(2)$, as mentioned by OP himself. One may check by inspection that the $X_n$ hit every element in $Y_n$, so that $X_n=Y_n$.

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  • $\begingroup$ @ Qmechanic, thanks for your wonderful comment. So do you mean that set $Y$ is always "larger" than set $X$ ? Is this conclusion also true for single spin-$\frac{1}{2}$ ? In spin-$\frac{1}{2}$ example, is $Y$ a group ? And is only the set $X$ the spin-rotation group for spin-$\frac{1}{2}$ while $Y$ not ? $\endgroup$
    – Kai Li
    May 5, 2013 at 17:33
  • $\begingroup$ @ Qmechanic, In spin-$\frac{1}{2}$ example, it's easy to show that $Y$ is a subset of group $SU(2)$(then a subset of $X$), if what you say "$X$ is a subset of $Y$" is simultaneously true, then we arrive at $X=Y=SU(2)$ for spin-$\frac{1}{2}$, right? $\endgroup$
    – Kai Li
    May 6, 2013 at 8:13
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    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 7, 2013 at 15:20
  • $\begingroup$ @K-boy: I would prefer to only answer mathematical question, and not to address semantical questions, which should anyway be easily to Google. In particular, I'm uncertain which connotation you associate with the words, such as, e.g., 'spin-rotation group', 'angular momentum' and 'spin'. $\endgroup$
    – Qmechanic
    May 11, 2013 at 21:37
  • $\begingroup$ Hi, I just realize that the conclusions in your answer seem not to be totally correct. Here may be a counterexample: Let $M_3=\sigma_z\otimes I +I\otimes\sigma_z, M_2=\sigma_y\otimes I +I\otimes\sigma_y$, where $\sigma_{y,z}$ are the Pauli matrices, and $I$ is $2\times2$ identity matrix. Then $X_{n=4}$ should be $SU(2)/Z_2=SO(3)$ rather than $SU(2)$ for $n=4$, which violates Eq.(3) in your answer. $\endgroup$
    – Kai Li
    Mar 3 at 21:19

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